|f 430 Moment of Resistance.
t;
1 p We may also represent the moment in terms of the limiting
!! stress/(sometimes/c, and sometimes/t). Then:
if Jj Bending moment = moment of resistance
j| Bm=/Z ........................ (4)
II
I/ and Z is known as the modulus of section,*
^
|| Let y = distance of furthest fibre from axis:
I By (i) /= by (3) Bm = — and by (4) Bm=/Z
Z-— .-. Z = *...... (5)
p P ° y
and Bm=/- .
_____I (Seej). 1079.)
The value of Z can now be found; thus,
p
1
% f
r i • rr, r " '" ^ f b frfi
rectangular Sections /Z =/------r - =/-—
and for circular sections /Z = /—— -r- -
----------------------------£---- J 64 2,
Graphic Solution for Moment of Resistance.—
Taking, first, a rectangular section (Fig. 384), draw the neutral
axis A B. Then c D will be the line of limiting (or greatest) stress,
and the value of any horizontal fibre E F to resist stress will be
found by projecting to c D and joining c D to N, thus obtaining
the intercept G M. Every fibre being thus treated, the sum of
the -virtual stress areas will be the areas c{# N and H j N, which
each make one force of the couple when multiplied by the
limiting stress f. K and L are the centres of gravity of the areas.
Moment of resistance (generally) = one force x total arm
Moment of rectangular section = f\area C D N [• x arm K L
Unsymmetrical sections are treated at Figs. 389, 390 by this
method, which can be applied to any section. (SeeApp.IL>p. 847.)
•* Z = virtual area x arm. (See Figs. 384, 385.)